everyone. I have a question regarding the convergence of a certain limit. I've been fiddling with it but its been proving quite evasive. What I am trying to calculate is the Grundwald-Letnikov derivative to an arbitrary order of a certain function centered at zero. I am using the derivative such that $e^x$ is a fix point, or the lowerlimit is at negative infinity.
I can state this more plainly without fractional calculus, so that my question is:
If $f(w) = \sum_{n=0}^\infty a_n w^n$ is an entire function of order zero, so for all $\rho > 0$ there exists a constant $C_\rho$ such that $|f(w)|\le C_\rho e^{|w|^{\rho}}$
If we define $$g_n(z) = n^z \sum_{k=0}^\infty (-1)^k \dbinom{z}{k} f(-\frac{k}{n})$$
does $g_n \to g$ converge uniformly or does the whole thing diverge? I know it works on some functions that grow even faster than $f$, however, I'm having trouble with $f$ itself
I'm thinking if we have some compact $\Omega$ such that $\forall z \in \Omega$ we get $|z|< R$, then, I've seen the inequality $\Big|\dbinom{z}{k}\Big| \le C \frac{R^k}{k!}$ and this gives me that:
$$|g_n(z)| \le C n^R \sum_{k=0}^\infty \frac{R^k}{k!} e^{\frac{k}{n}} = C n^R e^{R/e^{1/n}}$$
but this diverges as $n \to \infty$. However this should tell us that $g_n$ converges
Similarly I tried: $$|g_n(z)| \le C_p n^R \sum_{k=0}^\infty \frac{R(R+1)\cdots (R+k-1)}{k!}e^{(\frac{k}{n})^\rho}$$
but this series diverges for all $n$ with no hope of telling us anything.
And this is where I'm stuck. If anyone has any suggetions I'd be happy to hear them :). Do I need stronger conditions on $f$ to guarantee convergence? If so does anyone know what those conditions may be?
Thanks a lot, any help will be greatly appreciated.